Solve the inequality both algebraically and graphically. Draw a number line graph of the solution and give interval notation.

Question

5xplus1less thanminusStartFraction 2 x Over 5 EndFraction
plus5
Question content area bottom
Part 1
The solution as an inequality is
  
enter your response here.
​(Type an inequality. Simplify your answer. Use integers or fractions for any numbers in the​ expression.)
Part 2
The solution in interval notation is
  
enter your response here.
​(Type your answer in interval notation. Use integers or fractions for any numbers in the​ expression.)
Part 3
Choose the correct graph of the solution below.
A.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 27 EndFraction. The region to the right of a left bracket at approximately 0.8 is shaded.
0
1
0
B.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 27 EndFraction. The region to the left of a right bracket at approximately 0.8 is shaded.
0
1
0
C.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 27 EndFraction. The region to the left of a right parenthesis at approximately 0.8 is shaded.
0
1
0
D.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 27 EndFraction. The region to the right of a left parenthesis at StartFraction 20 Over 27 EndFraction is shaded.
0
1
0
E.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 27 EndFraction. The region to the left of a right bracket at StartFraction 20 Over 27 EndFraction is shaded.
0
1
0
F.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 27 EndFraction. The region to the right of a left parenthesis at approximately 0.8 is shaded.
0
1
0
G.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 27 EndFraction. The region to the right of a left bracket at StartFraction 20 Over 27 EndFraction is shaded.
0
1
0
H.
An infinite number line, labeled from 0 to 1, has tick marks in increments of StartFraction 1 Over 27 EndFraction. The region to the left of a right parenthesis at StartFraction 20 Over 27 EndFraction is shaded.
0
1
0

Answer 1

To solve the inequality \( 5x + 1 < -\frac{2x}{5} + 5 \), let's first simplify and isolate \( x \).

Part 1: Solve the Inequality Algebraically

1. Clear the fraction by multiplying the entire inequality by 5 (this keeps the inequality the same since 5 is positive): \[ 25x + 5 < -2x + 25 \]

2. Rearranging terms: \[ 25x + 2x < 25 - 5 \] \[ 27x < 20 \]

3. Dividing both sides by 27: \[ x < \frac{20}{27} \]

So the solution as an inequality is: \[ x < \frac{20}{27} \]

Part 2: Write the Solution in Interval Notation

The solution in interval notation is: \[ (-\infty, \frac{20}{27}) \]

Part 3: Choose the Correct Graph

Now, we need to select the correct graph to represent the solution. This inequality indicates that \( x \) can take any value less than \( \frac{20}{27} \) but not equal to \( \frac{20}{27} \).

Looking at the answer choices:

• The only choices that have shading to the left of \( \frac{20}{27} \) and that indicate \( \frac{20}{27} \) is not included (so a parenthesis instead of a bracket) are:

• C: "The region to the left of a right parenthesis at approximately \( \frac{20}{27} \) is shaded."

• H: "The region to the left of a right parenthesis at \( \frac{20}{27} \) is shaded."

Either C or H would be correct depending on how they represent \( \frac{20}{27} \).

Summary of Solutions

• Inequality Solution: \( x < \frac{20}{27} \)
• Interval Notation: \( (-\infty, \frac{20}{27}) \)
• Graph: Options C or H, depending on the representation of the number line.